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Assume $\{\alpha_1,\cdots\alpha_n\},\{\epsilon_1,\cdots,\epsilon_n\} $ are both orthonormal basis of Euclidean Space $V$. Consider the matrix $$A=(\cos\langle\alpha_i,\epsilon_j\rangle)_{n\times n}$$ $\langle\alpha_i,\epsilon_j\rangle$ denotes the angle between $\alpha_i,\epsilon_j$.
Prove 1. A is orthogonal. 2. Every orthogonal matrix can express as the form of $A$.

i.e prove $$\cos\langle\alpha_i,\epsilon_1\rangle\cdot \cos\langle\alpha_j,\epsilon_1\rangle+\cos\langle\alpha_i,\epsilon_2\rangle\cdot\cos\langle\alpha_j,\epsilon_2\rangle+\\ \cdots+\cos\langle\alpha_i,\epsilon_n\rangle\cdot\cos\langle\alpha_j,\epsilon_n\rangle=0$$ for $i\ne j$.

And

$$\cos\langle\alpha_i,\epsilon_1\rangle\cdot \cos\langle\alpha_j,\epsilon_1\rangle+\cos\langle\alpha_i,\epsilon_2\rangle\cdot\cos\langle\alpha_j,\epsilon_2\rangle+\\ \cdots+\cos\langle\alpha_i,\epsilon_n\rangle\cdot\cos\langle\alpha_j,\epsilon_n\rangle=1$$ for $i=j$.

How to deal with the sum?

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  • $\begingroup$ So... I guess $\cos<\alpha_i,\epsilon_j>$ is just a very convoluted way to express the usual meaning of $\langle \alpha_i,\epsilon_j\rangle$? $\endgroup$ – Saucy O'Path Oct 18 '18 at 7:23
  • $\begingroup$ @ Saucy O'Path It's the cosine of the angle between $\alpha_i,\epsilon_j$ $\endgroup$ – Jaqen Chou Oct 18 '18 at 7:25
  • $\begingroup$ Yes, "a very convoluted way..." et cetera. $\endgroup$ – Saucy O'Path Oct 18 '18 at 7:26
  • $\begingroup$ huh... what's the usual way... $\endgroup$ – Jaqen Chou Oct 18 '18 at 7:34
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Hint. The cosine of the angle between two unit vectors $u$ and $v$ is just their dot product, i.e. $u^Tv$. So, if $P$ is the augmented matrix containing the $\alpha_j$s as columns and $Q$ is the augmented matrix containing the $\varepsilon_j$s as columns, then $A=P^TQ$.

For the second part, take $Q=A$ with an appropriate $P$.

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  • $\begingroup$ So since $P,Q$ are orthogonal,$A^TA=Q^TPP^TQ=I $. Could you also give some hints about every orthogonal have form of this $A=P^TQ$? $\endgroup$ – Jaqen Chou Oct 18 '18 at 9:21
  • $\begingroup$ emm... I am trying to find out exactly $P$ and $Q$..Am I miss something... $\endgroup$ – Jaqen Chou Oct 18 '18 at 9:47
  • $\begingroup$ @JaqenChou Ok, see my new edit. $\endgroup$ – user1551 Oct 18 '18 at 9:48

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